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Alex17521 [72]
3 years ago
7

A voltaic cell is constructed with two silver-silver chloride electrodes, where the halfreaction is AgCl (s) + e- → Ag (s) + Cl-

(aq) E° = +0.222 V
The concentrations of chloride ion in the two compartments are 0.0222 M and 2.22 M, respectively. The cell emf is __________ V.

A) 0.212

B) 0.118

C) 0.00222

D) 22.2

E) 0.232
Chemistry
1 answer:
densk [106]3 years ago
6 0

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ag(s)+Cl^-(aq.)\rightarrow AgCl(s)+e^-

<u>Reduction half reaction (cathode):</u>  AgCl(s)+e^-\rightarrow Ag(s)+Cl^-(aq.)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}]_{diluted}}{[Cl^{-}]_{concentrated}}

where,

n = number of electrons in oxidation-reduction reaction = 1

E_{cell} = ?

[Cl^{-}]_{diluted} = 0.0222 M

[Cl^{-}]_{concentrated} = 2.22 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}

E_{cell}=0.118V

Hence, the cell potential of the cell is +0.118 V

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